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(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where is the mass of the sphere and is the radius of the sphere.

(b) Given the moment of inertia of a disc of mass and radius about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

(a) The moment of inertia (M.I.) of a sphere about its diameter =
                                  

According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. 

The M.I. about a tangent of the sphere = MR 

                                                  = 

(b) The moment of inertia of a disc about its diameter =  

According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body. 

The M.I. of the disc about its centre =  +   

                                               = 


The situation is shown in the given figure. 

                                   

Applying the theorem of parallel axes, 

The moment of inertia about an axis normal to the disc and passing through a point on its edge, 

 M.I =  + MR2 =
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Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

Let at a certain instant two particles be at points P and Q, as shown in the following figure.



Angular momentum of the system about point P, 

Lp = mv × 0 + mv × d 

    =  
mvd                                   ...(i) 

Angular momentum of the system about point Q, 

LQ = mv × d + mv × 0  

    =  
mvd                                   ...(ii) 


Consider a point R, which is at a distance y from point Q.

i.e.,               
QR = 

∴                   PR = d – y  

Angular momentum of the system about point R, 

LR = mv × (d - y) + mv × y 

    = mvd - mvy + mvy 

    = mvd                                   ...(iii) 

Comparing equations (i)(ii), and (iii), we get

LP = LQ = LR                              ...(iv)

From equation 
(iv), we infer that that the angular momentum of a system does not depend on the point about which it is taken.

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Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is rwith components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.



When the particle is confined to move in the x-y plane, the direction of angular momentum is along z direction. 

Hence, the result. 
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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.


The free body diagram of the bar is shown in the following figure.



Length of the bar, l = 2 m 

Tand T2 are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have

T1 Sin 36.90 = T2 Sin 53.1

       =  

⇒      T1 =  T2
For rotational equilibrium, on taking the torque about the centre of gravity, we have

      T1 (Cos 36.9) × d = T2 Cos 53.1 (2 - d) 

            T1 × 0.800 d = T2 × 0.600 (2 - d

   × T2 × 0.800d = T2 (0.600 × 2 - 0.600d

          1.067d + 0.6d = 1.2 

∴                         d = 1.2 / 1.67 

                             = 0.72 m 

Hence, the centre of gravity of the given bar lies 0.72 m from its left end. 

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A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Mass of the car, m = 1800 kg 

Distance between the front and back axles, d = 1.8 m

Distance between the centre of gravity and the back axle = 1.05 m 

The various forces acting on the car are shown in the following figure.


 


Rf and Rare the forces exerted by the level ground on the front and back wheels respectively. 

At translational equilibrium, 

Rf + Rb = mg 

            = 1800 × 9.8

            = 17640 N                               ... (i)

For rotational equilibrium, on taking the torque about the C.G., we have

R(1.05) = R(1.8 - 1.05)

        

          Rb = 1.4 Rf                                 ...(ii) 

Solving equations (i) and (ii), we get

1.4Rf + Rf = 17640 

            Rf = 7350 N 

∴          Rb = 17640 – 7350

                = 10290 N 

Therefore, the force exerted on each front wheel =   =  3675 N.
The force exerted on each back wheel =  =  5145 N 

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